Question

An electric bulb of 100 W - 300 V is connected with an AC supply of 500 V and $\frac{150}{\pi }$ Hz. The required inductance to save the electric bulb is

• A. $2H$
• B. $\frac{1}{2}H$
• C. $4H$
• D. $\frac{1}{4}H$

Answer 1

Answer: (C)

$4H$

The correct option is C
The voltage at which the bulb operates is $300v$ And the aac supply has voltage $500v$
Therefore, the difference of the voltage $V_1=500-300=200v$
Will appear the inductance of the spred by the coil of the bulb.
Therefore $V_l=\omega L\times I\Rightarrow \frac{P}{v}=\frac{100}{500}=\frac{1}{5}A$.
Thus, $V_l=2\pi fL\times I\Rightarrow 200=2\pi \frac{150}{\pi }l\frac{1}{5}\Rightarrow l=\frac{200}{60}=3.33\approx 4H$